# Special Factoring

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• February 29th 2008, 05:06 PM
Mad
Special Factoring
$(a - b)^3 - b^3$
$[(a - b) - b][(a - b)^2 + (a - b)b + b^2]$

Is that right?
• February 29th 2008, 05:15 PM
Soroban
Correct, Mad!

. . . Nice going!

• February 29th 2008, 05:50 PM
Mad
Gah, this one is driving me nuts :

Factor:

$27a^4 - a$
$a(27a^3 - 1)$

Or can I do more?
• February 29th 2008, 05:55 PM
mr fantastic
Quote:

Originally Posted by Mad
Gah, this one is driving me nuts :

Factor:

$27a^4 - a$
$a(27a^3 - 1)$

Or can I do more?

Note that $27a^3 - 1 = (3a)^3 - 1^3$, a difference of two perfect cubes ......
• February 29th 2008, 05:56 PM
Mad
I was thinking about that but didn't know what to do with the negative one. I should have known better, thanks. (Headbang)

EDIT: Just in case...

$(a)(27a - 1)(27a^2 + 27a + 1)$
• February 29th 2008, 06:16 PM
mr fantastic
Quote:

Originally Posted by Mad
I was thinking about that but didn't know what to do with the negative one. I should have known better, thanks. (Headbang)

EDIT: Just in case...

$(a)(27a - 1)(27a^2 + 27a + 1)$

Should be 3a's, NOT 27a's ......
• March 3rd 2008, 06:17 PM
Mad
Sorry, I don't understand. I looked back over my notes but I'm kind of confused. (Thinking)
• March 3rd 2008, 06:36 PM
topsquark
Quote:

Originally Posted by Mad
Gah, this one is driving me nuts :

Factor:

$27a^4 - a$
$a(27a^3 - 1)$

Or can I do more?

$27a^3 - 1 = (3a)^3 - (1)^3$

and
$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

So let $x = 3a$ and $y = 1$

Thus
$27a^3 - 1 = (3a)^3 - (1)^3 = (3a - 1)((3a)^2 + (3a)(1) + (1)^2)$

$= (3a - 1)(9a^2 + 3a + 1)$

-Dan
• March 3rd 2008, 06:37 PM
Mad
Oh, now I understand, thank you. (Smile)